Answer:
1.63ₓ10⁻⁶ g of U
139.03 g of H
0.385 g of O
141.8 g of Pb
Explanation:
In first place, we need to convert the number of atoms to moles, as we know that 1 mol of anything occupies 6.02×10²³ particles
Therefore:
4.12×10¹⁵ atoms of U . 1 mol / 6.02×10²³ atoms = 6.84×10⁻⁹ moles of U
8.37×10²⁵ atoms of H . 1 mol /6.02×10²³ atoms = 139.03 moles of H
1.45×10²² atoms of O . 1 mol /6.02×10²³ atoms = 0.0241 moles of O
4.12×10²³ atoms of Pb . 1 mol /6.02×10²³ atoms = 0.684 moles of Pb
Moles . Molar mass = Mass (g)
6.84×10⁻⁹ moles of U . 238.03 g/mol = 1.63ₓ10⁻⁶ g of U
139.03 moles of H . 1 g/mol = 139.03 g of H
0.0241 moles of O . 16 g/mol = 0.385 g of O
0.684 moles of Pb . 207.2 g/mol = 141.8 g of Pb
Explanation:
<em><u>Solutions. 1. If 47 g of KCl dissolved in enough water to give 375 mL of soloution, what is the molarity ... vo volume of solute . ... v/v ethanol, how much 95% v/v ethanol ... prepare 200. mL ...</u></em>
<u>Answer:</u> The molality of the solution is 0.1 m.
<u>Explanation:</u>
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute = 27.1 g
= Molar mass of solute = 27.1 g/mol
= Mass of solvent = 100 g
Putting values in above equation, we get:
Hence, the molality of the solution is 0.1 m.
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
Answer:Effect of Catalysts on the Activation Energy. Catalysts provide a new reaction pathway in which a lower Activation energy is offered. A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller energy barrier.
Explanation:
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